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3 years ago

12

Why can polar water molecules pass through a cell membrane?

1 answer:

TiliK225 [7]3 years ago

5 0

<span> Through osmosis because the membrane is partially permeable, osmosis is the movement of water molecules from a region of lower solute concentration to a higher solute concentration.
is the answer
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Please help! Question is on the bottom

choli [55]

It’s extremely bad quality I really can’t read it

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3 years ago

A total solar eclipse occurs because the moon can totally block the sun. If the moon was smaller or farther from earth, a total

Setler79 [48]

Answer:

both

Explanation:

id say that it could occur but also not as much. the moon would be smaller and further from the earth to where we would barely be able to see it. if the full moon is barely visible then im sure the total solar eclipse wouldn't be as noticeable as it is now. but thats just my opinion

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2 years ago

What is the mass of one mole of carbon-12?

velikii [3]

Answer:

Hello

12 grams

The mass of one mole of carbon-12 atoms is 12 grams.

Hope it helps You.....

Explanation:

4 0

3 years ago

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Experiment on distillation

Temka [501]

Answer:

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2 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

2 years ago